${\left[Fe{\left(CN\right)}_6\right]}^{3-}$ has a value of only 1.74 BM?

Ans/.       $As we know {\mu }_m=\sqrt{n(n+2}BM$
$Where, \mu =magnetic moment$
$it \mu =number of unpaired electrons$
$and \mu =1.74 i.e., n=1$
$\mu =5.92 i.e., n=5$

${\left[Fe{\left(CN\right)}_6\right]}^{3-}$involves $d^{2}s{p}^3$hybridisation with one unpaired electron (as shown by its

magnetic moment 1.74 BM) and ${\left[Fe{\left(H_{2}0\right)}_6\right]}^{3+}$involves sp${}^3{d}^2$hybridisation with five

unpaired electrons (because magnetic moment equal to 6.92 BM)

CN${}^{-}$ is stronger ligand than H${}_2$,O according to specttochemical series.${∆}_0>P$ for CN${}^{-}$hence,

fourth electron will pair itself, Whereas for water pairing will not happen for ${\left[Fe{\left(CN\right)}_6\right]}^{3-}$ the

electronic configuration of Fe${}^{3+}$ is

One unpaired electron 

For ${\left[Fe{(H}_{}\right]}^{}$₂O)63+ the electronic configuration of Fe${}^{3+}$ is 

Five unpaired electron 

Hence, ${\left[Fe{\left(CN\right)}_6\right]}^{3-}$ and ${\left[Fe{\left(H_{2}O\right)}_6\right]}^{3+}$ are inner obital an douter orbital complex respectively